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sergiy2304 [10]
3 years ago
11

Write this trinomial in factored form 2a^2 +7a+3

Mathematics
1 answer:
EleoNora [17]3 years ago
8 0

Answer:

(2a + 1)(a + 3) = 2a² + 7a + 3

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Find the value of x. 3x(5+2)=16
Iteru [2.4K]

Answer:

16/21

Step-by-step explanation:

We are given the equation 3x(5 + 2) = 16.

By order of operations, let's add the two numbers in the parentheses first:

3x(5 + 2) = 16

3x(7) = 16

Now, we must multiply 3x by 7:

3x * 7 = 16

21x = 16

Divide both sides by 21:

x = 16/21

The answer is thus 16/21.

<em>~ an aesthetics lover</em>

8 0
3 years ago
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Which shows 2^-2 * 2^6 in exponential form
kompoz [17]

Answer:

16

Step-by-step explanation:

2^-2=0.25

2^6=64

times

=16

5 0
3 years ago
In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 3.0 m/s up a 22.0° inclined track
Aloiza [94]

Answer:

Approximately 0.31\; \rm m, assuming that g = 9.81\; \rm N \cdot kg^{-1}.

Step-by-step explanation:

Initial kinetic energy of the sled and its passenger:

\begin{aligned}\text{KE} &= \frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2} \times 14\; \rm kg \times (3.0\; \rm m\cdot s^{-1})^{2} \\ &= 63\; \rm J\end{aligned} .

Weight of the slide:

\begin{aligned}W &= m \cdot g \\ &= 14\; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \\ &\approx 137\; \rm N\end{aligned}.

Normal force between the sled and the slope:

\begin{aligned}F_{\rm N} &= W\cdot  \cos(22^{\circ}) \\ &\approx 137\; \rm N \times \cos(22^{\circ}) \\ &\approx 127\; \rm N\end{aligned}.

Calculate the kinetic friction between the sled and the slope:

\begin{aligned} f &= \mu_{k} \cdot F_{\rm N} \\ &\approx 0.20\times 127\; \rm N \\ &\approx 25.5\; \rm N\end{aligned}.

Assume that the sled and its passenger has reached a height of h meters relative to the base of the slope.

Gain in gravitational potential energy:

\begin{aligned}\text{GPE} &= m \cdot g \cdot (h\; {\rm m}) \\ &\approx 14\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times h\; {\rm m} \\ & \approx (137\, h)\; {\rm J} \end{aligned}.

Distance travelled along the slope:

\begin{aligned}x &= \frac{h}{\sin(22^{\circ})} \\ &\approx \frac{h\; \rm m}{0.375}\end{aligned}.

The energy lost to friction (same as the opposite of the amount of work that friction did on this sled) would be:

\begin{aligned} & - (-x)\, f \\ = \; & x \cdot f \\ \approx \; & \frac{h\; {\rm m}}{0.375}\times 25.5\; {\rm N} \\ \approx\; & (68.1\, h)\; {\rm J}\end{aligned}.

In other words, the sled and its passenger would have lost (approximately) ((137 + 68.1)\, h)\; {\rm J} of energy when it is at a height of h\; {\rm m}.

The initial amount of energy that the sled and its passenger possessed was \text{KE} = 63\; {\rm J}. All that much energy would have been converted when the sled is at its maximum height. Therefore, when h\; {\rm m} is the maximum height of the sled, the following equation would hold.

((137 + 68.1)\, h)\; {\rm J} = 63\; {\rm J}.

Solve for h:

(137 + 68.1)\, h = 63.

\begin{aligned} h &= \frac{63}{137 + 68.1} \approx 0.31\; \rm m\end{aligned}.

Therefore, the maximum height that this sled would reach would be approximately 0.31\; \rm m.

7 0
3 years ago
With full explanation from the internet like before<br> 3x2-6x+5=0
Elodia [21]

Answer:

\sf x=1+i\sqrt{\dfrac{2}{3}}   \ \quad and  \quad \:x=1-i\sqrt{\dfrac{2}{3}}

Explanation:

<u>Given Expression</u>:

  • 3x² - 6x + 5 = 0

Use the Quadratic Formula:

\sf x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}  \  \ when   \ \  ax^2 + bx + c = 0

<u>insert coefficients</u>

\Longrightarrow \sf x = \dfrac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:3\cdot \:5}}{2\cdot \:3}

\Longrightarrow \sf x = \dfrac{\left6\right\pm \sqrt{-24} }{6}

\Longrightarrow \sf x = \dfrac{\left6\right\pm 2\sqrt{6}i}{6}

\Longrightarrow \sf x =1 \pm   i\dfrac{\sqrt{6} }{3}

\Longrightarrow \sf x=1+i\sqrt{\dfrac{2}{3}},  \quad 1-i\sqrt{\dfrac{2}{3}}

6 0
2 years ago
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How long is a line of 1 million centimeter cubes be?
fenix001 [56]

Answer:

1m cm

10m mm

Step-by-step explanation:

4 0
3 years ago
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